Problem: Find $\lim_{x\to 1}\dfrac{{1-x^2}}{\ln(x)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-2$ (Choice B) B $-1$ (Choice C) C $0$ (Choice D) D The limit doesn't exist.
Solution: Substituting $x=1$ into $\dfrac{{1-x^2}}{\ln(x)}$ results in the indeterminate form $\dfrac{0}{0}$. Furthermore, as the expression involves mixed function types, it's not possible to manipulate it algebraically in a way that will help us find the limits. Therefore, we should use L'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to 1}\dfrac{{1-x^2}}{\ln(x)} \\\\ &=\lim_{x\to 1}\dfrac{\dfrac{d}{dx}[{1-x^2}]}{\dfrac{d}{dx}[\ln(x)]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to 1}\dfrac{(-2x)}{\left(\dfrac1x\right)} \\\\ &=\dfrac{-2(1)}{1} \gray{\text{Substitution}} \\\\ &=-2 \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to 1}\dfrac{\dfrac{d}{dx}[{1-x^2}]}{\dfrac{d}{dx}[\ln(x)]}$ actually exists. In conclusion, $\lim_{x\to 1}\dfrac{{1-x^2}}{\ln(x)}=-2$.